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The handle http://hdl.handle.net/1887/40676 holds various files of this Leiden University dissertation.
Author: Ciocanea Teodorescu, I.
Title: Algorithms for finite rings Issue Date: 2016-06-22
Chapter 5
A miscellaneous collection of algorithms
5.1 Testing if a ring is a field
One basic question we may ask ourselves when presented with a finite ring is if it is not in fact a field.
Lemma 5.1.1. Let p be a prime and R a finite commutative Fp-algebra. Then the following are equivalent:
(i) The map F : R → R given by x 7→ xp is injective.
(ii) R has no nonzero nilpotent elements.
(iii) R is a field.
Proof. Note thatF is an Fp-linear map.
(i)⇒(ii): Suppose there exists 0 6= x ∈ R and n ∈ Z>1 such that xn−1 6= 0 and xn= 0. Choosed ∈ Z≥0maximal such thatdp < n. Then (d + 1)p ≥ n and d + 1 < n, so 0 6=xd+1 ∈ ker(F ).
(ii)⇒(iii): IfR has no nonzero nilpotent elements, then R is semisimple (see The- orem 1.4.9, part (iii)). SinceR is commutative, it must be a field.
(iii)⇒(i) is clear.
Theorem 5.1.2. There exists a deterministic polynomial-time algorithm that, given a finite ringR, determines whether or not R is a field.
Proof. If char(R) is not a prime or R is not commutative, then R is not a field. This can be tested using Theorem 1.1.3 and Theorem 3.3.1. OtherwiseR is a commutative F-algebra and we use Lemma 5.1.1, part (i) to test whether it is a field.
Note 5.1.3. Another deterministic polynomial-time algorithm for testing if a finite ring is a field is given in [4], Section 5, Theorem 4.
65
Note 5.1.4. A deterministic polynomial-time algorithm for the case where R+ = (Z/pZ)n, for some primep and some n ∈ Z>0 is given in [23], Section 4.
5.2 Testing if a ring is simple
We have seen in Theorem 1.1.3 that primality testing is in P. It is therefore natural to ask whether we can construct a deterministic polynomial-time algorithm which decides whether a finite ring is simple.
Lemma 5.2.1. Let R be a semisimple ring. Then the centre of R is a field if and only ifR is simple.
Proof. SinceR is semisimple, it is a finite product of simple rings. Hence the centre ofR is the product of the centres of these simple rings, and so, is a finite product of fields. This product is then a field itself if and only ifR was simple to begin with.
Theorem 5.2.2. There exists a deterministic polynomial-time algorithm that, given a finite ring R, determines whether R is simple, and if it is, outputs a prime p and two positive integersm and n such that R ∼= Mn(Fpm).
Proof. Computek := Z(R) and test if it is a field using Theorem 5.1.2. If k is not a field, thenR cannot be simple. If it is a field, then R is a finite-dimensional algebra over k, and we can compute its Jacobson radical using Theorem 3.4.1. If J(R) 6= 0, thenR cannot be simple. If J(R) = 0, then R is a semisimple algebra whose centre is a field, which by Lemma 5.2.1 impliesR is simple. Proceed by computing m ∈ Z>0, the dimension of k over Fp, andn ∈ Z>0, the size of the matrix ring, both of which can be done in polynomial time.
Note 5.2.3. Given two finite rings R and R0, we can now decide if they are simple and isomorphic: simply compare the size of the matrices which will be produced by Theorem 5.2.2 and test if the centres ofR and R0are isomorphic fields, using Theorem 3.5.2.
The algorithm above does not explicitly exhibit an isomorphism between R and Mn(Z(R)). We can use Theorem 3.5.2 to get an isomorphism of fields Z(R) ∼= Fpm, for somem ∈ Z>0, but we can say no more than that.
The problem of exhibiting an isomorphism R ∼= Mn(Z(R)) is often referred to as the explicit isomorphism problem, and has received recent attention in [45, 46]. In the case that Z(R) is a finite field and n is a power of 2, this problem has a deterministic polynomial-time solution. In general, the problem of finding an isomorphism between finite algebras over finite fields is not believed to be NP-hard, but is at least as hard as the graph isomorphism problem ([41, 52]).
Iuliana Cioc˘anea-Teodorescu 67
5.3 Testing if a module is simple
Testing simplicity of a module can also be done in polynomial time.
Theorem 5.3.1. There exists a deterministic polynomial-time algorithm that, given a finite ringR and a finite R-module M , determines whether M is simple or not.
Proof. By Schur’s Lemma (Theorem 1.6.1), ifM is simple, then EndR(M ) is a division ring, so we start by computingn := char(EndR(M )) = exp(M+) using Theorem 3.3.1.
Ifn is not a prime, then we conclude that M is not simple. Otherwise, if n is a prime, thennR is a two-sided ideal in R, and R0 :=R/nR is an algebra over a finite field, so we may compute its Jacobson radical using Theorem 3.4.1. NowM is an R0-module, sincenR annihilates M .
If J(R0) does not annihilateM , then M is not simple over R0, and henceM is not simple overR. Otherwise, M is an R0/ J(R0)-module and by Theorem 1.6.4, it is now enough to test whether EndR0/ J(R0)(M ) is a field, which can be done by Theorem 5.1.2.
Second proof. Alternatively, compute
I := ann(M ) = ker(R → EndZ(M+), r 7→ r · m),
where “·” denotes the action ofR on M . Then M is a faithful R/I-module and so if M is simple, we claim that R/I is simple as a ring. To see this, suppose M is simple.
Then the Jacobson radical ofR/I annihilates M , but since M is faithful, J(R/I) = 0, henceR/I is semisimple. Now M is a faithful simple module over a semisimple ring, soR/I must in fact be simple.
We thus begin by testing simplicity ofR/I as a ring, using Theorem 5.2.2. If R/I is not simple, then M cannot be simple and we are done. Otherwise, the algorithm in Theorem 5.2.2 will output a prime p, and two integers m, n such that R/I ∼= Mn(Fpm) as rings. Now, by Theorem 1.4.2, the only simple Mn(Fpm)-module, up to isomorphism, is (Fpm)n, and the other modules over Mn(Fpm) are direct sums of (Fpm)n. Moreover,R/I-modules are exactly the R-modules annihilated by I. Hence M is simple over R if and only if |M | = pnm.
5.4 Testing if a module is projective
For many future algorithms it will be very useful to be able to test if a module is projective.
Theorem 5.4.1. There exists a deterministic polynomial-time algorithm that, given a finite ring R and a R-module M , together with a generating set of cardinality d, for some d ∈ Z≥0, determines if M is R-projective or not, and if it is, produces a splitting of the natural surjectionRd M .
Proof. Recall thatM is projective if and only if the natural surjection f : Rd M has a left inverse. The latter can be tested using Proposition 2.5.1, which will also produce a left inverse.
Second proof. Another way to determine whetherM is projective comes as a conse- quence of Theorem 4.1.1, sinceM is projective if and only if M is a direct summand of Rd. We compute the largest isomorphic common direct summand of Rd and M , say S. If M ∼=S, then M is projective and the isomorphism M → S, which is also produced by the algorithm, induces a splitting ofRd→ M . Otherwise the algorithms concludes thatM is not projective.
5.5 Constructing projective covers
Recall the definition of a projective cover given in Definition 1.6.24 and the fact that over left-artinian rings, all modules have a projective cover, unique up to isomorphism (Theorem 1.6.25).
Theorem 5.5.1. There exists a deterministic polynomial-time algorithm that, given a finite ringR and a finite R-module M , outputs a projective cover of M .
Proof. Use the algorithm in the proof of Theorem 4.1.3 to construct a sequence of R-modules (Si)ti=1, two sequences of integers (ai)ti=1 and (ci)ti=1, and a two-sided nilpotent ideal a such that
R/a ∼=
t
M
i=1
Siai and M/aM ∼=
t
M
i=1
Sici,
where for all 1 ≤ i ≤ t, we have that ai > 0 and ci ≥ 0. Relabel, to write R/a ∼= L
i∈ISi and M/aM ∼= L
j∈JSj. Then for each j ∈ J we have a surjective map gj :M Sj, such that aM =T
j∈Jker(gj).
For each j ∈ J, pick j0 ∈ I such that Sj ∼= Sj0. Since Sj0 is a direct summand of R/a, there exists an idempotent ej ∈ R/a such that Sj0 ∼= (R/a)ej (see Theorem 1.5.4). To findej, look at the image of 1 under the isomorphismR/a ∼=L
i∈ISi and identify the entry corresponding toj0. Replace all other entries by zeros and letejbe the preimage of this element under the same isomorphism.
Thus, the decomposition R/a ∼=L
i∈ISi gives rise to a sequence of idempotent elementse1, . . . , e|J|in R/a such that for all j ∈ J we have Sj0 ∼= (R/a)ej.
By Proposition 1.5.8, these idempotents can then be lifted deterministically in polynomial time to idempotentse1, . . . , e|J| in R. For all j ∈ J, let Pj =Rej. Since ej is an idempotent in R, we can write R = Rej ⊕ R(1 − ej), so Pj is projective.
Hence we can construct a sequence of mapsfj such that for eachj ∈ J the following diagram commutes:
Pj
M Sj,
πj
fj
gj
where πj : Pj → Sj is the natural projection map. This can be done by solving a system of linear equations overR. Let P =L
j∈JPj and letf : P → M be the direct sum of thefj.
Iuliana Cioc˘anea-Teodorescu 69
We claim that for eachj ∈ J, the pair (Pj, πj) is a projective cover ofSj. Clearly πjis surjective andPjis projective. We need to show that ker(πj) ⊆sPj. LetN ≤ Pj
be a submodule such that ker(πj) +N = Pj. By construction, ker(πj) = aPj. Since a ⊆ J(R), by Nakayama’s Lemma we must have N = Pj. Since taking projective covers commutes with direct sums, (P,L
j∈Jπj) is a projective cover for M/aM . Since a is nilpotent,f is surjective and (P, f ) is a projective cover of M .
5.6 Constructing injective hulls
Recall the definition of injective hulls given in Definition 1.6.28 and the fact that in- jective hulls exist for modules over any ring. Moreover, two injective hulls of a module M are isomorphic (Theorems 1.6.29).
To construct injective hulls, we will make use of the character module (see Defi- nition 1.9.1). Recall that for a finite ring R, the character functor defines a duality betweenfgRMand MfgR. (see Theorem 1.9.2). Moreover, the following holds.
Proposition 5.6.1. Let R be a left-noetherian ring and M a finitely generated R- module. Then
(i) M is projective inRMif and only if M is projective infgRM.
(ii) M is injective inRMif and only if M is injective infgRM.
Proof. The “only if” directions of both (i) and (ii) are clear. We prove the converse statements below.
SupposeM is projective inRM. ThenM is a direct summand of a finitely gener- ated freeR-module, so it is projective infgRM.
Suppose M is injective in RM. Since R is left-noetherian, all its left ideals are finitely generated. Hence by Theorem 1.6.12, part (iii) (Baer’s test),M is injective in
fg RM.
Note 5.6.2. The left-noetherian condition on R is not needed for part (i).
Corollary 5.6.3. Let R be a finite ring and M an R-module. Then M is injective overR if and only if cM is projective over Ro, the opposite ring of R.
Note 5.6.4. IfR is a finite ring and M is a finite R-module, then we may take
M = Homc Z(M,1
eZ/Z), (5.1)
wheree is a multiple of exp(M+).
Theorem 5.6.5. There exists a deterministic polynomial-time algorithm that, given a finite ringR and a finite R-module M , computes an injective hull of M .
Proof. Compute cM = HomZ(M,1eZ/Z), where e is a multiple of exp(M+). Using Theorem 5.5.1, compute a projective cover (P, f ) of cM over Ro. Now set I := bP = HomZ(P,e10Z/Z), where e0 is a multiple of exp(P+). Then (I, g) is an injective hull of M by Theorem 1.9.2. The algorithm also produces a map g : cM ,→ I, given byc precomposition withf , such that im(g) ⊇eI.
5.7 Testing if a module is injective
Recall that a module is injective if and only if it is isomorphic to its injective hull (Theorem 1.6.30).
Theorem 5.7.1. There exists a deterministic polynomial-time algorithm that, given a finite ringR and a finite R-module M , determines if M is injective.
Proof. Construct an injective hullI of M using Theorem 5.6.5. Now check if the map g : M ,→ I produced by Theorem 5.6.5 is bijective.
5.8 Testing if a ring is quasi-Frobenius
Recall that a finite ringR is quasi-Frobenius if R is left self-injective (Theorem 1.7.1, Definition 1.7.2).
Theorem 5.8.1. There exists a deterministic polynomial-time algorithm that, given a finite ringR, determines whether R is quasi-Frobenius.
Proof. Use Theorem 5.7.1 to determine if the left-regularR-module RR is injective.
5.9 Constructive tests for existence of injective and surjective module homomorphisms
In this section we discuss the algorithmic problem of testing for existence and of finding injective and surjective homomorphisms between two finite length modules over a ring R. If R is a finite-dimensional algebra over a field, this problem can be cast in the context of matrix completion, and has been shown to be NP-hard in [42].
In view of the results of [10, 15] and of Theorem 4.1.2, this result is striking. It is not however an isolated type of result: the subgraph isomorphism problem is an NP-hard problem, while the graph isomorphism problem is believed to be NP-intermediate.
While in the general case, testing constructively for existence of injective and sur- jective module homomorphisms is NP-hard, with certain restrictions on the modules considered, the problem turns out to be tractable. We are interested in the case where R is a finite ring and one of the modules is either projective or injective over R, for which the problem simplifies somewhat.
Iuliana Cioc˘anea-Teodorescu 71
Theorem 5.9.1. There exists a deterministic polynomial-time algorithm that, given a finite ringR and two finite R-modules M and N , one of which is projective, deter- mines whether there exists a surjection M N. If one exists, the algorithm exhibits one such.
Proof. If N is projective, then it suffices to test whether N is a direct summand of M , which can be done by Theorem 4.1.3. This will also produce a surjection M N.
IfM is projective, then we proceed by constructing a projective cover (P, f ) of N . Note that existence of a surjection M N is equivalent to existence of a surjection M P . If there exists a surjection M P , simply compose it with f to get a surjectionM N. Conversely, if there exists a surjection M N, then, since P is a projective cover ofN , there exists a surjective map g : M P making the following diagram commute
P
M N.
g f
This reduces the problem to the previous case.
Second proof for the case whereM is projective. If M is projective, we can also de- cide existence of a surjection M → N in a more direct manner. Use the algorithm in the proof of Theorem 4.1.3 to construct a two-sided nilpotent ideal a ⊂R and a sequence of R-modules (Si)ti=1 that is “compatible” both with M and with N , i.e.
such that
M/aM ∼=
t
M
i=1
Siai and N/aN ∼=
t
M
i=1
Sibi,
for someai, bi∈ Z≥0. This is done by running the algorithm in the proof of Theorem 4.1.3 on the ringR and the module M , and including N as one of the candidates for the isomorphism classes of simpleR-modules in the UPDATE subroutine. Note that, by construction, the algorithm ensures that for alli 6= j, we have HomR(Si, Sj) = 0.
We claim that existence of a surjectionM/aM N/aN, which can be easily tested by comparing ai and bi for each 1 ≤i ≤ t, is equivalent to existence of a surjection M → N . If there exists a surjection M → N , then clearly it induces a surjection M/aM N/aN. Conversely, if there exists a surjection f : M/aM N/aN, then, sinceM is projective, there exists a map f : M → N making the following diagram commute:
M N
M/aM N/aN,
f
f
andf is surjective since a is nilpotent.
Note 5.9.2. The case where M ∼=Rn, for some n ∈ Z>0 can be settled using the algorithm for computing the minimum number of generators of a module, given in Theorem 4.1.3.
Dually to Theorem 5.9.1, we have the following result:
Theorem 5.9.3. There exists a deterministic polynomial-time algorithm that, given a finite ringR and two finite R-modules M and N , one of which is injective, determines whether or not there exists an injectionM ,→ N . If one exists, the algorithm exhibits one such.
Proof. Letk := 1eZ/Z, where e ∈ Z>0 is a multiple of exp(M+) and exp(N+), and apply Theorem 5.9.1 to modules Homk(N, k) and Homk(M, k).
The remaining cases, not treated by Theorems 5.9.1 and 5.9.3, are constructive tests for existence of the followingR-module homomorphisms:
• P ,→ M , for P a projective module,
• M ,→ P , for P a projective module, and their respective duals,
• N I, for I an injective module,
• I N, for I an injective module.
Mimicking the construction given in the proof of Theorem 1.2 of [42], we settle these as being NP-hard, even when R is a finite local commutative ring and P = R.
This is done by a reduction from an instance of the nonsingular matrix completion problem, which is known to be NP-hard.
The nonsingular matrix completion problem is an algorithmic question that can be formulated as follows: given a square matrix A, whose entries are homogeneous linear polynomials in F[x1, . . . , xn], for some field F, decide if there exist values from F that can be assigned to the variables x1, . . . , xn, so as to makeA nonsingular. The constructive version of this problem asks for values ofx1, . . . , xnmakingA nonsingular to be exhibited.
Nonsingular matrix completion problems arise naturally in spaces of linear trans- formations. Letn ∈ Z>0 and F be a field. Let A ⊂ Mn(F) be a linear subspace and let {A1, . . . , Am} be a basis of A over F. Deciding existence of (resp. finding) a non- singular matrix inA is equivalent to deciding existence of (resp. finding) a sequence c1, . . . , cm∈ F such that Pm
i=1ciAi is nonsingular.
The complexity of the nonsingular matrix completion problem is very much de- pendent on the size of the field F (see [40, 42]). If F is “large enough”, then the Schwartz-Zippel lemma (see [81, 89]) provides an efficient randomized solution. How- ever, over finite fields, nonsingular matrix completion is NP-complete ([14, 40]).
Theorem 5.9.4. There exists a deterministic polynomial-time reduction from the decision (resp. constructive) version of nonsingular matrix completion to the problem of deciding existence of (resp. finding) an injective module homomorphism from a finite commutative local ringR containing a field, to an R-module M .
Iuliana Cioc˘anea-Teodorescu 73
Proof. Let F be a finite field and let U, V be two finite-dimensional F-vector spaces of the same dimension. Let 0 6=L ≤ HomF(U, V ) be a linear subspace. Consider the ring
R = F ⊕ U,
with componentwise addition and multiplication given by (a, x)(b, y) = (ab, ay + bx).
ThenR is a commutative local ring, with maximal ideal U , and U2= 0.
Put
M = L ⊕ V.
We makeM into an R-module by defining an action:
(a, u) · (l, v) := (al, av + l(u)), for alla ∈ F, u ∈ U, l ∈ L and v ∈ V .
Note that any homomorphism R → M is determined by the image of 1R. Let ψ : R → M be an R-module homomorphism, and suppose 1 7→ (l, v), for some (l, v) ∈ M . Then
im(ψ) = F(l, v) + (0, lU)
andψ ∈ HomR(R, M ) is injective if and only if l ∈ L is an isomorphism.
Theorem 5.9.5. There exists a deterministic polynomial-time reduction from the decision (resp. constructive) version of nonsingular matrix completion to the problem of deciding existence of (resp. finding) an injective module homomorphism from an R-module M to R, where R is a finite commutative local ring containing a field.
Proof. Let F be a finite field and let U, V be two finite-dimensional F-vector spaces of the same dimension. Let 0 6=L ⊆ HomF(U, V ) be a linear subspace. Consider the ring
R = F ⊕ L ⊕ U ⊕ V, with componentwise addition and multiplication given by
(f, l, u, v) · (f0, l0, u0, v0) = (f f0, f l0+f0l, f u0+f0u, f v0+f v + l(u0) +l0(u)).
ThenR is a commutative ring with unique maximal ideal L ⊕ U ⊕ V . Note that L ⊕ V is a two-sided ideal inR. Put
M := R/(L ⊕ V ) ∼= F ⊕ U.
Note that U2 = 0, so M also has the structure of a local commutative ring, with maximal idealU .
Note that
HomR(M, R) ∼= annR(L ⊕ V ) = L ⊕ U0⊕ V, (5.2)
whereU0=T
f∈Lker(f ) and the isomorphism in (5.2) is given by mapping f 7→ f (1).
Letφ be an R-module homomorphism. Then φ corresponds uniquely to an element (0, l0, u0, v0) ∈L ⊕ U0⊕ V and
im(φ) ∼= F(0, l0, u0, v0) + (0, 0, 0, l0U ).
Henceφ ∈ HomR(M, R) is injective if and only if l0∈ L is an isomorphism.
We consider now another weaker variant of the problem of testing constructively for injective and surjective module homomorphisms. Suppose we are given a finite ringR and two modules M and N . Instead of looking for a surjection M N, we may ask if there is an integer k such that there exists a surjective homomorphism f : Mk N . If such a pair (k, f ) exists, we would like to exhibit it. Note that we do not ask for k to be minimal with this property. This problem turns out to have an easy solution.
Theorem 5.9.6. There exists a deterministic polynomial-time algorithm that, given a finite ring R and two finite R-modules M, N , decides if there exists a pair (k, f ), wherek ∈ Z≥0 andf : Mk→ N is a surjective R-module homomorphism. If it exists, the algorithm exhibits such a pair.
Proof. Compute a set S of Z-generators of HomR(M, N ). If N =P
f∈Sf (M ), then output (|S|,P
f∈Sf ). Otherwise conclude that there does not exist a pair as required.
Dually, we have the following result:
Theorem 5.9.7. There exists a deterministic polynomial-time algorithm that, given a finite ring R and two finite R-modules M, N , decides if there exists a pair (k, f ), wherek ∈ Z≥0 andf : M → Nk is an injectiveR-module homomorphism. If it exists, the algorithm exhibits such a pair.
Proof. Compute a setS of Z-generators of HomR(M, N ). IfT
f∈Sker(f ) = {0}, then output (|S|,Q
f∈Sf ). Otherwise conclude that there does not exist a pair as required.
